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| − | A model $\mathfrak B$ is ''Jonsson'' if $|{\mathfrak B}|>\aleph_0$ and for every ${\mathfrak A}\prec {\mathfrak B}$, if $|{\mathfrak A}|=|{\mathfrak B}|$, then ${\mathfrak A}={\mathfrak B}$.
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| − | Gaifman <cite> gaifman1976:models </cite> and Knight <cite> knight1976:hanf </cite> independently showed that there are Jonsson models of $PA$.
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| − | Jonsson models $M$ of $PA$ of cardinality $\aleph_1$ are either $\aleph_1-like$ or are ''short''. A model $M$ of $PA$ is short there is an $a\in M$ such that the Skolem closure of $a$ is cofinal in $M$. Each known Jonsson model realizes uncountably many complete types.
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| − | Kossak has asked: Is there an $\aleph_1$-like Jonsson model $M\models PA$ such that $|\{{\rm tp}(a): a\in M\}|=\aleph_0$?
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| − | If $M\models PA$ is $\aleph_1$-like and recursively saturated, then $|\{{\rm tp}(a): a\in M\}|=\aleph_0$, but $M$ is not Jonsson. Therefore, another related question is: Is there a ''weakly Jonsson model'' $M\models PA$, i.e. a recursively saturated model $M\models PA$ such that for every recursively saturated $K\prec M$, if $|K|=|M|$, then $K=M$? The problem was posed in <cite> kossak1995:four </cite>.
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| − | {{References}}
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